$\textbf{Question:}$ In a $k$-connected graph $(k\ge2)$, any $k$ vertices lie on a common cycle.
$\textbf{Proof:}$ Let $S$ be a given set of $k$ vertices and consider a cycle $C$ with the maximum number of vertices from $S$. Suppose that some $v \in S - C$. Then by Menger theorem, there are $kv - C$ paths. Partition $C$ into at most $k - 1$ paths $P_j$, where the $i$th path begins from the $i$th vertex of $S$ on $C$ (in clockwise order say), and ends just before the $i + 1st$ vertex. Since $|S| = k$, by pigeonhole, two of the $v - C$ paths have their endpoints in the same $P_i$. Detouring along these two paths yields a cycle with more vertices of $S$, contradiction.
$\textbf{A follow-up question:}$ Show that if a graph is 2-connected, then every two cycles of maximum length have at least two common vertices.
Helpful comments are needed to prove this follow up.